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<p>Substituting (<a href="" class="xref" data-knowl="./knowl/eq5_1.html" title="Equation 5.2.1">(5.2.1)</a>) into the ODE, we have</p>
<div class="displaymath process-math" data-contains-math-knowls="./knowl/eq5_1.html ./knowl/eq5_2.html">
\begin{equation*}
\begin{aligned}
&amp;\sum_{n=0}^{\infty} n a_n x^{n-1}-\sum_{n=0}^{\infty} a_n x^n=0,\\
&amp;\to \sum_{k=-1}^{\infty} (k+1) a_{k+1} x^k-\sum_{n=0}^{\infty} a_n x^n=0,\\
&amp;\to \sum_{n=0}^{\infty} (n+1) a_{n+1} x^n-\sum_{n=0}^{\infty} a_n x^n=0,\\
&amp;\to \sum_{n=0}^{\infty} [(n+1) a_{n+1}-a_n] x^n=0.\\
\end{aligned}
\end{equation*}
</div>
<p class="continuation">If a power series is equal to zero, then all the coefficients must vanish. So we have</p>
<div class="displaymath process-math" data-contains-math-knowls="./knowl/eq5_1.html ./knowl/eq5_2.html">
\begin{equation}
\begin{aligned}
(n+1) a_{n+1}=a_n \to a_{n+1}=\frac{a_n}{n+1}.
\end{aligned}\tag{5.2.2}
\end{equation}
</div>
<p class="continuation">Equation (<a href="" class="xref" data-knowl="./knowl/eq5_2.html" title="Equation 5.2.2">(5.2.2)</a>) is called the <dfn class="terminology">recurrence relation</dfn>. From this relation, we have</p>
<div class="displaymath process-math" data-contains-math-knowls="./knowl/eq5_1.html ./knowl/eq5_2.html">
\begin{equation*}
\begin{aligned}
&amp;n=0: a_1=a_0,\\
&amp;n=1: a_2=\frac{a_1}{2}=\frac{a_0}{1\cdot 2}=\frac{a_0}{2!},\\
&amp;n=2: a_3=\frac{a_2}{3}=\frac{a_0}{3!},\\
&amp;\cdots\\
&amp;a_n=\frac{a_0}{n!},
\end{aligned}
\end{equation*}
</div>
<p class="continuation">where <span class="process-math">\(a_0\)</span> is an arbitrary constant. Thus, the solution is</p>
<div class="displaymath process-math" data-contains-math-knowls="./knowl/eq5_1.html ./knowl/eq5_2.html">
\begin{equation*}
y=\sum_{n=0}^{\infty} a_n x^n=a_0 \sum_{n=0}^{\infty} \frac{x^n}{n!}=a_0 e^x.
\end{equation*}
</div>
<span class="incontext"><a href="sec5_2.html#p-206" class="internal">in-context</a></span>
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